Logarithms and logarithmic functions

1. Introduction To
Logarithms

2. Why is logarithmic scale used to measure sound?

3. Ourfirstquestionthen
:
mustbe
?
Whatisalogarithm

4. -4 -3 -2 -1 1 2 3 4 5
-30
-20
-10
10
20
30
x
yx
y2=
One one Function

5. -8-7-6-5-4-3-2-1 12345678
-8
-6
-4
-2
2
4
6
8
x
y
(3,8)
(2,4)
(1,2)
(-1,1/2)
(-2,1/4)
(8,3)
(4,2)
(2,1)
(1/2,-1)
(-1/4,-2
)
x
y2=
Inverse of
x
y2=

6. The inverse of is the function
x
y2=
xy
2
log=

7. -8-7-6-5-4-3-2-1 12345678
-8
-6
-4
-2
2
4
6
8
x
y
(3,8)
(2,4)
(1,2)
(-1,1/2)
(-2,1/4)
x
y2=
)8log,8(
2
)4log,4(
2
)2log,2(
2
)2/1log,2/1(
2
)4/1log,4/1(
2
xy
2
log=

8. Domain of logarithmic function = Range of
exponential function =
Range of logarithmic function = Domain of
exponential function =

9. 1. The x-intercept of the graph is 1. There is
no y-intercept.
2. The y-axis is a vertical asymptote of the
graph.
3. A logarithmic function is decreasing if
0 < a < 1 and increasing if a > 1.
4. The graph contains the points (1,0) and
(a,1).
Properties of the Graph of a Logarithmic Function

10. Logarithmic Abbreviations
log
10 x = log x (Common log)
log
e x = ln x (Natural log)
e = 2.71828...

11. Of course logarithms have

a precise mathematical

definition just like all terms
. ’
in mathematics So lets
.
start with that

12. DefinitionofLogarithm
>0 1,
Supposeb andb≠
‘ ’
thereisanumber p
:
suchthat
log
bn=p if and only if b
p
=n

14. ,
The first and perhaps

the most important
,
step in understanding

logarithms is to realize

that they always relate

back to exponential
.
equations

15. You must be able to

convert an exponential

equation into logarithmic
.
form and vice versa
’
So lets get a lot of practice
!
with this

16. 1:
Example
:
Solution
log
28=3
:
We read this as ”the
2 8
log base of is
3 .
equal to ”
3
Write 2 8in logarithmic form.=

17. Example
1 :
a
Write 4
2
=16 in logarithmic form.
:
Solution
log
416=2
:
Read as “the
4 16
log base of
2 .
is equal to ”

18. 1 :
Example b
:
Solution
Write 2
-3
=
1
8
in logarithmic form.
log
2
1
8
=-3
1
Read as: "the log base 2 of is equal to -3".
8

19. , ’
Okay so now its

time for you to try
.
some on your own
1. Write 7
2
=49 in logarithmic form.
7
Solution: log49 2=

20. log
51=0:
Solution
2. Write 5
0
=1 in logarithmic form.

21. 3. Write 10
-2
=
1
100
in logarithmic form.
:
Solution
log
10
1
100
=-2

22. :
Solution
log
164=
1
2
4. Finally, write 16
1
2
=4
in logarithmic form.

23. Itisalsoveryimportanttobe

abletostartwithalogarithmic

expressionandchangethis
.
intoexponentialform

Thisissimplythereverseof
.
whatwejustdid

24. 1:
Example
Write log
381=4 in exponential form
:
Solution
3
4
=81

25. 2:
Example
Write log
2
1
8
=-3 in exponential form.
:
Solution
2
-3
=
1
8

26. ,
Okay now you try
.
these next three
1. Write log
10100=2 in exponential form.
3. Write log
273=
1
3
in exponential form.
2. Write log
5
1
125
=-3 in exponential form.

27. 1. Write log
10100=2 in exponential form.
:
Solution
10
2
=100

28. 2. Write log
5
1
125
=-3 in exponential form.
:
Solution
31
5
125
-
=

29. 3. Write log
273=
1
3
in exponential form.
:
Solution
27
1
3
=3

30. We nowknowthata logarithmis

perhaps bestunderstood

as being

closelyrelatedtoan
.
exponentialequation
,
Infact wheneverwe getstuck

inthe problems thatfollow

we willreturnto
.
this one simple insight

We mightevenstate a
. simple rule

31. ,
Whenworkingwithlogarithms
,
ifeveryouget“stuck” try

rewritingthe problemin
.
exponentialform
,
Conversely whenworking
,
withexponentialexpressions
,
ifeveryouget“stuck” try

rewritingthe problem
.
inlogarithmicform

32. ’
Letsseeifthissimple
rule

canhelpussolvesome

ofthefollowing
.
problems

33. :
Solution
’
Lets rewrite the

problem in
.
exponential form
6
2
=x
’ !
We re finished
6
Solve for x: log 2x=
Example 1

34. :
Solution
5
y
=
1
25

Rewrite the

problem in
.
exponential form
Since
1
25
=5
-2æ
è
ç
ö
ø
÷
5
y
=5
-2
y=-2
5
1
Solve for y: log
25
y=
Example 2

35. 3
Example
Evaluate log
327.

Try setting this up like
:
this
:
Solution
log
327=y
Now rewrite in exponential
.
form
3
y
=27
3
y
=3
3
y=3

36. These next two

problems tend to be

some of the trickiest
.
to evaluate
,
Actually they are

merely identities and

the use of our
simple

rule
.
will show this

37. Example
4
Evaluate: log
77
2
:
Solution

Now take it out of the logarithmic

form
.
and write it in exponential form
log
77
2
=y, .
First we write the problem with a variable
7
y
=7
2
y=2

38. Example
5
Evaluate: 4
log416
:
Solution
4
log416
=y, .
First we write the problem with a variable
log
4y=log
416

Now take it out of the exponential

form
.
and write it in logarithmic form
Just like 2
3
=8 converts to log
28=3
y=16

39. Ask your

teacher about

the last two
.
examples

They may show

you a nice
.
shortcut

40. ,
Finally we wanttotake a lookat

the PropertyofEqualityfor
.
LogarithmicFunctions
Suppose b>0 and b¹1.
Then log
bx
1=log
bx
2 if and only if x
1=x
2
, ,
Basically with logarithmic functions

ifthe bases match on both sides ofthe equal
, .
sign then simplysetthe arguments equal

41. Logarithmic Equations

42. Example
1
Solve: log
3(4x+10)=log
3(x+1)
:
Solution
‘3’
Since the bases are both we
.
simply set the arguments equal
4x+10=x+1
3x+10=1
3x=-9
x=-3

43. Example
2
Solve: log
8(x
2
-14)=log
8(5x)
:
Solution
‘8’
Since the bases are both we simply set the
.
arguments equal
x
2
-14=5x
x
2
-5x-14=0
(x-7)(x+2)=0
Factor
(x-7)=0 or (x+2)=0
x=7 or x=-2
continued on the next
page

44. 2
Example
continued
Solve: log
8(x
2
-14)=log
8(5x)
:
Solution
x=7 or x=-2
2
It appears that we have solutions
.
here

If we take a closer look at the
,
definition of a logarithm however

we will see that not only must we
,
use positive bases but also we see

that the arguments must be
. -2
positive as well Therefore is not
. a solution
’
Let s end this lesson by taking a .
closer look at this

45. Ourfinalconcern thenis to

determine whylogarithms like
.
the one beloware undefined

Cananyone give
?
us anexplanation
2
log(8)-

46. One easy explanation is to simply

rewrite this logarithm in
.
exponential form
’
We ll then see why a negative
.
value is not permitted
, .
First we write the problem with a variable
2
y
=-8

Now take it out of the logarithmic

form
.
and write it in exponential form
2 -8 ?
What power of would gives us
2
3
=8 and 2
-3
=
1
8

Hence expressions of this type are
.
undefined
2
log( 8) undefined WHY?- =
2
log ( 8)y- =

49. Thatconcludes ourintroduction
.
tologarithms Inthe lessons to

followwe willlearnsome important
.
properties oflogarithms

One ofthese properties willgive

us a veryimportanttool
which

we needtosolve exponential
. ’
equations Untilthenlets

practice withthe basicthemes
.
ofthis lesson

50. That’s All Folks !